Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: Rimi on February 13, 2012, 11:40:32 PM
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The air inside a refrigerator (volume 100 L) is cooled from 20 °C to 0 °C after closing the door.
What is the force needed to open the refrigerator door (area 1 m2)? The refrigerator is supposed
to be air-tight.
I really have no clue how to approach this problem. When the temperature decreases the volume also decreases..ok. I know Force = Pressure x Area, but I don't think that will help at all.
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Volume is constant, so it is the pressure that goes down.
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Use the ideal gas law to calculate the pressure after cooling, and then use the formula you provided.
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That's what I figured, but what is the initial pressure? That's what I'm not understanding. Unless my answer is supposed to be in terms of pressure?
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Start with 1 atm.
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If you've opened the door to put something in, the pressure inside the refrigerator is the same as the pressure outside the refrigerator - 1 atm. Cooling doesn't start (in your example) until the door is closed, so the initial pressure when you begin cooling is also 1 atm.
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Thanks I got it. :)
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would you like to share your results
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would you like to share your results
Wouldn't that be against the rules if someone came by with the same question?
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Wouldn't that be against the rules if someone came by with the same question?
I will have to ask Socrates - philosophy was not my strong suit.
;D
I am assuming your class is done with this question and has been graded on for you and your classmates.
by the way
We often check peoples results to let them know if they did it right so long as they show all work.
You can send me a private message to help my curiosity if you want
;)
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would you like to share your results
Wouldn't that be against the rules if someone came by with the same question?
Oh no, you are allowed to answer your own questions - and it makes us feel good to know that you got the right answer :D
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Since this has been posted for a while and the homework has probably been graded, I thought I would attempt to solve this as a exercise in honing my skills.
With the variables
P as pressure
V as volume
n as amount of substance (moles)
T as temperature
R as the universal gas constant
using the ideal gas law formula
PV=nRT
rearrange
PV/nT=R
for before refrigerator is closed
P1V1/n1T1=R
for after refrigerator is closed and cooled
P2V2/n2T2=R
since R is a constant
P1V1/n1T1=
P2V2/n2T2
since the volume before and after are the same
since the moles before and after are the same
P1/T1=P2/T2
multiplying both sides by T2
P2=P1T2/T1
Given
Force = Pressure x Area
Door area = 1 m2
V1 = V2 = 100 L
T1 = 293.15K (20C) T2 = 273.15K (0C)
assumed
P1 = 1 atm (Atmosphere)
looked up
1 atm = 101325 pascal (Pa)
Then
P2= ((101325Pa)(273.15K))/(293.15K)
P2= 94412.156745693331059184717721303
rounding
P2= 94412 Pa
Thus
outside the refrigerator is P1 = 101325 Pa
inside the refrigerator is P2 = 94412 Pa
The difference between inside and outside pressure = 6913 Pa
recollect the given
area of the refrigerator door is 1 m2
force = pressure x area
looked up
1 Pa = 1 N/m2 where N is newton and m is meter
Since the door is 1m2 the force to open = 6913 N
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I was really surprised by the answer, which for me was in kilograms, completely forgetting that they are not an 'acceptable' unit of force.
I did remember that doors have hinges.
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Force = Pressure x Area,
Can you calculate the pressure difference between the inside and the outside of the refrigerator? That would be the pressure you would have to overcome to open the door, and the force required would be related to that pressure and the area of the door.
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@fledarmus
I wrote
outside the refrigerator is P1 = 101325 Pa
inside the refrigerator is P2 = 94412 Pa
The difference between inside and outside pressure = 6913 Pa
later
1 Pa = 1 N/m2 where N is newton and m is meter
Since the door is 1m2 the force to open = 6913 N
Alternatively I could have said
outside the refrigerator is P1 = 101325 Pa
inside the refrigerator is P2 = 94412 Pa
and
1 Pa = 1 N/m2 where N is newton and m is meter
Since the door is 1m2
the force outside the refrigerator is 101325 N
the force inside the refrigerator is 94412 N
The difference between these forces is 6913 N
which is the force necessary to open the door
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@fledarmus
Is the second way a better way of presenting the results,since the final answer in both is the same?
Or
is my physics rusty
@Wastrel
For me it would have been useful to couch what the force would be like by stating in lifting of grams against gravity. I do not have a since in my mind about newtons.
Example
The newtons are for opening the door feel the same as lifting a ??? gram weight against gravity.
do you want to fill in the grams?
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Sorry Billnotgatez, for some reason all that showed for me this morning was the first post. I thought somebody else was starting a new thread! Completely disregard my response.
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@fledarmus
<I smack my forehead>
I understand you might have missed what I posted
But did I get the right answer
<I smack my forehead again>
:) ;) ;D
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I just did a back of envelope calculation. I looked up atmospheric pressure in kilograms per square meter (10332) and multiplied by 20/293 (as T and P are proportional). 705 Kg tallies with billnotgatez and I'm happy with this as the force against the door.
The force holding the door shut is not the same as the force needed to open it. Doors have hinges. Making the not unrealistic assumption that the handle is on the opposite edge to the hinge then the force needed is half the force keeping the door shut. My answer is then 353Kg. Astonishing for what seems to be mild and reasonable conditions.
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353 kg or
778 lb and 3.70 oz
seems a bit much to open a refrigerator door
but when I divide by 10 (~gravity acceleration 9.80665)
I get 691.3 kg which halved is 345.65 kg
or
762 lb and 0.44 oz
I would have never thought I was that much of a muscle man.
;D
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Fridge door is never really air tight, so in reality the pressure inside is equal to that on the outside.